Ask question

Ask question

All categories

3 years ago

6

A runner is participating in the Boston Marathon. He has run 12 miles of the 26.2 mile course. What percent of the course has he

run so far?

2 answers:

Pani-rosa [81]3 years ago

6 0

45.8% or rounded up to 46%

Natasha2012 [34]3 years ago

4 0

46% just so you know i rounded this up!

You might be interested in

Please help me on this question

Vesnalui [34]

Jill is picking up everyone's waterbill so she is paying $60 plus $125 (being half of the water bill) so that comes out to $185.
D=185 in this case

Ron is paying for 1/3 of the rent plus the other half of the water bill. That comes out to 1/3R+125.
A=1/3R+125

And then Jack is paying for Jill's rent as well as his own. That is 2/3 of the rent.
C=2/3R

and then the choice not used is the 'no match' option. Hope this helps :)

7 0

3 years ago

(⚠please help⚠)Which languages do many East Africans speak in addition to their many Native languages?

Klio2033 [76]

Answer:

Step-by-step explanation:

A

4 0

3 years ago

Read 2 more answers

The manager of a fleet of automobiles is testing two brands of radial tires and assigns one tire of each brand at random to the

Darya [45]

Answer:

Brand 1 Brand 2 Difference

37734 35202 2532

45299 41635 3664

36240 35500 740

32100 31950 150

37210 38015 −805

48360 47800 560

38200 37810 390

33500 33215 285

Sum of difference = 2532+ 3664+740+150 −805+ 560 +390 +285 = 7516

Mean = $d=\frac{7516}{8}$

Mean = $d=939.5$

a) d= 939.5

$\text{Sample Standard deviation, s} = \sqrt{\dfrac{(x-\bar{x})^2}{n-1}}$

$=\sqrt{\dfrac{(2532-939.5)^2+(3664-939.5)^2+(740-939.5)^2 ...+(285-939.5)^2}{8-1}}$

=1441.21

b)SD= 1441.21

c)Calculate a 99% two-sided confidence interval on the difference in mean life.

confidence level =99%

significance level =α= 0.01

Degree of freedom = n-1 = 8-1 =7

So, $t_{\frac{\alpha}{2}}=3.499$

Formula for confidence interval $= \left( \bar{X} \pm t_{\frac{\alpha}{2}} \times \frac{s}{\sqrt{n}} \right)$

Substitute the values

confidence interval $= 939.5 \pm 3.499 \times \frac{1441.21}{\sqrt{8}} \right)$

confidence interval $= 939.5 - 3.499 \times \frac{1441.21}{\sqrt{8}} \right)$ to $= 939.5 + 3.499 \times \frac{1441.21}{\sqrt{8}} \right)$

Confidence interval $−843.396\$ to $2722.396$

8 0

3 years ago

A manufacturer has 6,305 yards of material to make

docker41 [41]

Answer:

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

G(x) = -2x^3 – 15x^2 + 36x

shusha [124]

Consider the function $G(x) = -2x^3 - 15x^2 + 36x.$ First, factor it:

$G(x) = -2x^3 - 15x^2 + 36x=-x(2x^2+15x-36)=\\ \\=-x\cdot 2\cdot \left(x-\dfrac{-15-\sqrt{513}}{4}\right)\cdot \left(x-\dfrac{-15+\sqrt{513}}{4}\right).$

The x-intercepts are at points $\left(\dfrac{-15-\sqrt{513} }{4},0\right),\ (0,0),\ \left(\dfrac{-15+\sqrt{513} }{4},0\right).$

1. From the attached graph you can see that

function is positive for $x\in \left(-\infrty, \dfrac{-15-\sqrt{513} }{4}\right)\cup \left(0,\dfrac{-15+\sqrt{513} }{4}\right);$
function is negative for $x\in \left(\dfrac{-15-\sqrt{513} }{4},0\right)\cup \left(\dfrac{-15+\sqrt{513} }{4},\infty\right).$

2. Since

$G(-x) = -2(-x)^3 - 15(-x)^2 + 36(-x)=2x^3-15x^2-36x\neq G(x)\ \text{and }\neq -G(x)$ the function is neither even nor odd.

3. The domain is $x\in (-\infty,\infty),$ the range is $y\in (-\infty,\infty).$

8 0

3 years ago

Read 2 more answers